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Quantity of water in 100kg of fresh fruits = ([latex]\frac {68} {100}[/latex] x 100) )kg Quantity of pulp in it = (100 - 68)kg = 32 kg Let the dry fruit be x kg Water in it = ([latex]\frac {20} {100}[/latex] x 100) )kg = [latex]\frac {x} {5}[/latex] kg Quantity of pulp in it = (x - [latex]\frac {x} {5}[/latex])kg = [latex]\frac {4x} {5}[/latex] kg Therefore, [latex]\frac {4x} {5}[/latex] kg = 32 => x = [latex]\frac {160} {4}[/latex] = 40 kg

Alcohol in 9 ml lotion = ([latex]\frac {50} {100}[/latex] × 9)ml = 4.5 ml Water in it = (9 – 4.5)ml = 4.5 ml Let x ml of water be added to it, then [latex]\frac {4.5} {9+x}[/latex] × 100 = 30 => [latex]\frac {4.5} {9+x}[/latex] = [latex]\frac {30} {100}[/latex] = [latex]\frac {3} {10}[/latex] => 3(9+x) = 45 => 27 + 3x = 45 => 3x = 18 => x = 6 Water to be added = 6 ml

Let A = set of students who play football and B = set of students play cricket. Then n(A) = 40, n (B) = 50 and n(A U B) = (100 - 18) = 82 n(A U B) = n(A) + n(B) – n(A ∩ B) n(A∩B) = n(A) + n(B) – n(AUB) = (40 + 50 -82) = 8 Percentage of the students who play both = 8%

Three consecutive numbers can be taken as (P - 1), P, (P + 1). So, (P - 1) + P + (P + 1) = 102 3P = 102 => P = 34. The lowest of the three = (P - 1) = 34 - 1 = 33.

Any two digit number can be written as (10P + Q), where P is the digit in the tens place and Q is the digit in the units place. P + Q = 10 ----- (1) (10Q + P) - (10P + Q) = 54 9(Q - P) = 54 (Q - P) = 6 ----- (2) Solve (1) and (2) P = 2 and Q = 8 The required number is = 28

Three consecutive even numbers (2P - 2), 2P, (2P + 2). (2P - 2) + 2P + (2P + 2) = 42 6P = 42 => P = 7. The middle number is: 2P = 14.

A + B = 60, A = 2B 2B + B = 60 => B = 20 then A = 40. 5 years, their ages will be 45 and 25. Sum of their ages = 45 + 25 = 70.

Their present ages be 3X and 4X. 5 years age, the ratio of their ages was 5:7, then (3X - 5):(4X - 5) = 5:7 X = 35 - 25 => X = 10. Their present ages are: 30, 40.

Present age of the son be P, he was born P years ago. The age of the man was: (48 - P). His age when the son was born should be equal to [latex]\frac {2} {3}[/latex] of P. (48 - P) = [latex]\frac {2} {3}[/latex] P 5P = 144 => P = 28.8

Let P's age and Q's age be 6x and 7x years respectively. Then, 7x - 6x = 4 => x = 4 Required ratio = (6x + 4) : (7x + 4) 28 : 32 = 7:8

Let the present ages of P and Q be 5x and 7x years respectively. Then, 7x - (5x + 6) = 2 2x = 8 => x = 4 Required sum = 5x + 7x = 12x = 48 years.

Let the present ages of X and Y be 5x and 6x years respectively. Then, [latex]\frac {(5x + 7)} {(6x + 7)}[/latex]= [latex]\frac {6} {7}[/latex] 7(5x + 7) = 6(6x + 7) => x = 7 X's present age = 5x = 35 years.

Their present ages be 3X and 4X. Let the present ages of the two brothers be x and 2x years respectively. Then, [latex]\frac {(x - 5)} {(2x - 5)}[/latex] = [latex]\frac {1} {3}[/latex] 3(x - 5) = (2x - 5) => x = 10 Required ratio = (x + 5) : (2x + 5) = 15 : 25 = 3:5